ABCD దీర్ఘచతురస్రంలో, ADE అనేది 18 సె౦.మీ2 వైశాల్యం కలిగిన సమద్విబాహు త్రిభుజం. అదేవిధంగా, EC = 2 × DE. ఒకవేళ AD = DE అయితే ABCD యొక్క వైశాల్యం ఎంత?

F1 S.C 6.2.20 Pallavi D3

  1. 72 సె౦.మీ2
  2. 90 సె౦.మీ2
  3. 108 సె౦.మీ2
  4. 126 సె౦.మీ2

Answer (Detailed Solution Below)

Option 3 : 108 సె౦.మీ2
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Detailed Solution

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⇒ ADE అనేది AD = DE = a అనే సమద్విబాహు త్రిభుజం.

ADE యొక్క ⇒ వైశాల్యం = 1/2 × AD × DE = 18

∴ 1/2 × a2 = 18

⇒ a = 6 సె౦.మీ

⇒ AD = DE = 6 సె౦.మీ

⇒ EC = 2 × DE = 2 × 6 = 12 సె౦.మీ

⇒ ∴ ABCD యొక్క  వైశాల్యం = AD × DC = 6 × (DE + EC) = 6 × (6 + 12) = 108 సె౦.మీ2

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